Multiply monogenic orders

Multiply monogenic orders

ATTILAB ´ERCZES, JAN-HENDRIKEVERTSE ANDK ´ALMAN´ GYORY˝

Abstract. LetA=Z[x1, . . . ,xr] Zbe a domain which is finitely generated overZand integrally closed in its quotient field L. Further, let K be a finite extension field ofL. AnA-order inK is a domainO Awith quotient fieldK which is integral overA. A-orders in Kof the typeA[↵]are called monogenic.

It was proved by Gy˝ory [10] that for any given A-orderO in K there are at most finitely manyA-equivalence classes of↵2 OwithA[↵] =O, where two elements↵, ofOare called A-equivalent if = u↵+afor someu 2 A^⇤, a2 A. If the number ofA-equivalence classes of↵withA[↵] =Ois at leastk, we callOktimes monogenic.

In this paper we study orders which are more than one time monogenic.

Our first main result is that ifK is any finite extension ofLof degree 3, then there are only finitely many three times monogenic A-orders in K. Next, we define two special types of two times monogenic A-orders, and show that there are extensionsK which have infinitely many orders of these types. Then under certain conditions imposed on the Galois group of the normal closure ofKover L, we prove thatK has only finitely many two times monogenicA-orders which are not of these types. Some immediate applications to canonical number systems are also mentioned.

Mathematics Subject Classification (2010):11R99 (primary); 11D99, 11J99 (secondary).

1. Introduction

In this introduction we present our results in the special case A= Z. Our general results over arbitrary finitely generated domains Aare stated in the next section.

LetK be an algebraic number field of degreed 2 with ring of integersOK. The number field K is calledmonogenicifOK = Z[↵]for some↵ 2 OK. This is equivalent to the fact that{1,↵, . . . ,↵^{d 1}}forms aZ-module basis forOK. The The research was supported in part by grants T67580 and T75566 (A.B., K.G.) of the Hungarian National Foundation for Scientific Research, and the J´anos Bolyai Research Scholarship (A.B.).

The work is supported by the T ´AMOP 4.2.1./B-09/1/KONV-2010-0007 project. The project is implemented through the New Hungary Development Plan, co-financed by the European Social Fund and the European Regional Development Fund. (A.B.)

Received July 18, 2011; accepted August 22, 2011.

existence of such a basis, called power integral basis, considerably facilitates the calculations inOK and the study of arithmetical properties ofOK.

The quadratic and cyclotomic number fields are monogenic, but this is not the case in general. Dedekind [4] gave the first example for a non-monogenic number field.

More generally, an orderOⁱⁿ K, that is a subring ofOK with quotient field equal to K, is said to be monogenic ifO = Z[↵] for some ↵ 2 O^{. Then for}

= ±↵+awitha 2 Zwe also haveO = Z[ ]. Such elements ↵, ofO^are calledZ-equivalent.

In this paper, we deal with the “Diophantine equation”

Z[↵] =O ^in↵2O ^(1.1)

where O is a given order in K. As has been explained above, the solutions of (1.1) can be divided into Z-equivalence classes. It was proved by Gy˝ory [7–9]

that there are only finitely manyZ-equivalence classes of↵ 2 Owith (1.1), and that a full system of representatives for these classes can be determined effectively.

Evertse and Gy˝ory [5] gave a uniform and explicit upper bound, depending only ond = [K : Q], for the number ofZ-equivalence classes of such↵. For various generalizations and effective versions, we refer to Gy˝ory [13].

In what follows, the following definition will be useful.

Definition. An orderO^{is calledk}times monogenic,if there are at leastkdistinct Z-equivalence classes of↵with (1.1), in other words, if there are at leastkpairwise Z-inequivalent elements↵₁, . . . ,↵_k 2O^{such that}

O=Z[↵₁] = · · · =Z[↵_k].

Similarly, the order O ^{is called}precisely/at mostk times monogenic, if there are precisely/at mostkZ-equivalence classes of↵with (1.1).

It is not difficult to show that any orderOin a quadratic number field is precisely one time monogenic, i.e., there exist ↵ 2 O with (1.1), and these ↵ are allZ^- equivalent to one another.

Our first result is as follows.

Theorem 1.1. Let K be a number field of degree 3. Then there are at most finitely many three times monogenic orders inK.

This result is a refinement of work of B´erczes [1].

The bound 3 is best possible,i.e., there are number fields K having infinitely many two times monogenic orders. We believe that ifKis an arbitrary number field of degree 3, then with at most finitely many exceptions, all two times monogenic orders in K are of a special structure. Below, we state a theorem which confirms this if we impose some restrictions onK.

Let K be a number field of degree at least 3. An order O ^{in K} is called of type Iif there are↵, 2O^{and a}a¹3^aa²4 2GL(2,Z^{)such that}

K =Q^(↵), O=Z[↵] =Z[ ], = a1↵+a2

a₃↵+a₄, a36=0. (1.2)

Notice that is notZ-equivalent to↵, sincea36=0 andK has degree at least 3. So orders of type I are two times monogenic.

OrdersOof type II exist only for number fields of degree 4. An orderO^{in a} quartic number field K is called oftype IIif there are↵, 2O^anda0,a₁,a₂,b₀, b₁,b₂2Z^witha0b₀6=0 such that

K =Q^(↵), O=Z[↵] =Z[ ], (1.3)

=a₀↵²+a₁↵+a₂, ↵=b₀ ²+b₁ +b₂.

Orders of type II are certainly two times monogenic. At the end of this section, we give examples of number fields having infinitely many orders of type I, respec- tively II.

LetEbe a field of characteristic 0, andF =E(✓)/Ea finite field extension of degreed. Denote by✓⁽¹⁾, . . . ,✓^(d)the conjugates of✓overE, and byGthe normal closureE(✓⁽¹⁾, . . . ,✓^(d))ofFoverE. We callF mtimes transitiveoverE(m d) if for any two orderedm-tuples of distinct indices(i1, . . . ,im),(j1, . . . ,jm)from {1, . . . ,d}, there is 2Gal(G/E)such that

(✓⁽ⁱ¹⁾)=✓^(j1), . . . , (✓^(im))=✓^(jm). If E=Q, we simply say thatFismtimes transitive.

We denote bySnthe permutation group onnelements.

Our result on two times monogenic orders is as follows.

Theorem 1.2. (i)Let Kbe a cubic number field. Then every two times monogenic order in Kis of type I.

(ii)LetK be a quartic number field of which the normal closure has Galois group S4. Then there are at most finitely many two times monogenic orders in K which are not of type I or of type II.

(iii)LetK be a four times transitive number field of degree at least5. Then there at most finitely many two times monogenic orders inKwhich are not of type I.

In Section 2 we present some immediate applications of our results to canonical number systems. In Section 3 we formulate generalizations of Theorems 1.1 and 1.2 for the case that the ground ring is an arbitrary integrally closed domain which is finitely generated over Z. Sections 4–6 contain auxiliary results, and Sections 7–9 contain our proofs.

Our proofs of Theorems 1.1 and 1.2 use finiteness results on unit equations in more than two unknowns, together with some combinatorial arguments. At present, it is not known how to make the results on unit equations effective, therefore we are not able to determine effectively the three times monogenic orders in Theorem 1.1, or the two times monogenic orders not of type I or II in Theorem 1.2. Although it is possible to estimate from above the number of solutions of unit equations, it is because of the combinatorial arguments in our proofs that we are not able to estimate from above the numbers of exceptional orders in Theorems 1.1 and 1.2.

We finish this introduction with constructing number fields having infinitely many orders of type I, respectively type II.

LetKbe a number field of degree 3 which is not a totally complex quadratic extension of a totally real field. By Dirichlet’s Unit Theorem, for any proper sub- field L of K, the rank of O^⇤_L (the group of units of the ring of integers of L) is smaller than that ofO^⇤_K. We show thatK has infinitely many orders of type I. Take

a1a2

a3a4 2GL(2,Z^)witha36=0. Suppose that there isu02O^⇤_K ^{such thatu}0⌘a4

(moda3). This is the case for instance ifa3 = 1. By the Euler-Fermat Theorem for number fields, there is a positive integert such thatu^t ⌘1 (moda3)for every u2O^⇤K. Hence the group of unitsu2O^⇤K withu⌘1 (moda3)has finite index in O^⇤K. Consequently, there are infinitely many unitsu2O^⇤K withu⌘a₄ (moda₃).

By our assumption onK, among these, there are infinitely manyuwithQ^(u)= K. For each suchu, put

↵:=u a4

a3

, :=a1↵+a2

a3↵+a4

.

Then clearly, K =Q(↵). From the minimal polynomial ofuwe derive a relation u ¹ = f(u)with f 2 Z[X]. Hence = (a1↵+a2)f(a3↵+a4) 2Z[↵]. Since

= (a4 a2)/( a3 +a1) and a3 +a1 = ±u ¹, we obtain in a similar fashion↵ 2Z[ ]. Therefore,Z[↵] = Z[ ]. By varying ^a_a¹₃^a_a²₄ anduwe obtain infinitely many orders of type I inK.

For instance, foru2O^⇤K we haveZ[u] =Z[u ¹]and the discriminant of this order is the discriminant of (the minimal polynomial of)u. By Gy˝ory [8, Corollaire 2.2], there are at most finitely many units u 2 O^⇤_K of given discriminant. Hence there are infinitely many distinct orders amongZ[u](u2O^⇤_K^).

We now construct quartic fields with infinitely many orders of type II. The construction is based on the theory of cubic resolvents, see van der Waerden [20, Section 64].

Letr,s be integers such that the polynomial f(X) = (X² r)² X s is irreducible and has Galois groupS4. There are infinitely many such pairs(r,s)(see, e.g., Kappe and Warren [14]). Denote by↵⁽¹⁾ = ↵,↵⁽²⁾,↵⁽³⁾,↵⁽⁴⁾ the roots of f and let K :=Q(↵). Define

⌘₁ := (↵⁽¹⁾+↵⁽²⁾)(↵⁽³⁾+↵⁽⁴⁾)=(↵⁽¹⁾+↵⁽²⁾)²,

⌘₂ := (↵⁽¹⁾+↵⁽³⁾)(↵⁽²⁾+↵⁽⁴⁾)=(↵⁽¹⁾+↵⁽³⁾)²,

⌘₃ := (↵⁽¹⁾+↵⁽⁴⁾)(↵⁽²⁾+↵⁽³⁾)=(↵⁽¹⁾+↵⁽⁴⁾)².

Then

(X ⌘₁)(X ⌘₂)(X ⌘₃)= X³ 4r X²+4s X 1. (1.4) Take p⌘₁=↵⁽¹⁾+↵⁽²⁾, p⌘₂=↵⁽¹⁾+↵⁽³⁾, p⌘₃=↵⁽¹⁾+↵⁽⁴⁾.

Then p

⌘₁·p

⌘₂·p

⌘₃=1. (1.5)

By the Gauss-Fermat Theorem over number fields, there exists a positive integert such that

⌘₁^t ⌘1 (mod 4). (1.6)

Consider form =0,1,2, . . .the numbers

↵_m:= ¹₂⇣p⌘₁^1+2mt +p⌘₂^1+2mt +p⌘₃^1+2mt⌘ ,

m:= ¹₂⇣p⌘₁ ^{1 2mt} +p⌘₂ ^{1 2mt} +p⌘₃ ^{1 2mt}⌘ .

The numbers ↵_m are invariant under any automorphism that permutes↵⁽²⁾,↵⁽³⁾,

↵⁽⁴⁾,i.e., under any automorphism that leavesKinvariant, hence they belong toK. Further, they have four distinct conjugates, soQ^(↵m)= K. Next, by (1.5),

m=↵²_m rm, ↵_m= m² sm, where

rm = ¹₄⇣

⌘₁^1+2mt+⌘¹₂^+2mt+⌘¹₃^+2mt⌘ , s_m = ¹₄

⇣

⌘₁^{1 2mt}+⌘₂^{1 2mt}+⌘₃^{1 2mt}⌘ .

By (1.4),(1.6),r_m,s_m are rational integers, hence↵_m, _mare algebraic integers for everym. We thus obtain for every non-negative integerman orderZ[↵_m] =Z[ m] of type II inK.

We claim that among the ordersZ[↵_m]there are infinitely many distinct ones.

Denote by Dmthe discriminant ofZ[↵_m]. ThenDmis equal to the discriminant of

↵_m, and a straightforward computation shows that this is equal to the discriminant of⌘₁^1+2mt. By [8, Corollaire 2.2], we have|Dm|! 1asm ! 1. This implies our claim.

2. Application to canonical number systems

LetKbe an algebraic number field of degree 2, andOan order inK. A nonzero element↵inOis called abasis of a canonical number system(or CNS basis) for Oif every nonzero element ofOcan be represented in the form

a₀+a₁↵+ · · · +a_m↵^m

with m 0, ai 2 {0,1, . . . ,|NK/Q(↵)| 1} for i = 0, . . . ,m, and am 6= 0.

Canonical number systems can be viewed as natural generalizations of radix repre- sentations of rational integers to algebraic integers.

When there exists a canonical number system inO^{, then}O is called a CNS order. Orders of this kind have been intensively investigated; we refer to the survey paper [2] and the references given there.

It was proved by Kov´acs [15] and Kov´acs and Peth˝o [16] thatOis a CNS order if and only ifO is monogenic. More precisely, if↵ is a CNS basis inO^{, then it} is easily seen that O = Z[↵]. Conversely, O = Z[↵]does not imply in general that↵is a CNS basis. However, in this case there are infinitely many↵⁰which are Z-equivalent to ↵ such that↵⁰ is a CNS basis for O. A characterization of CNS bases inOis given in [16].

The close connection between elements↵ofO^withO=Z[↵]and CNS bases inOenables one to apply results concerning monogenic orders to CNS orders and CNS bases. The results presented in Section 1 have immediate applications of this type. For example, it follows that up toZ-equivalence there are only finitely many canonical number systems inO^.

We say that O ^{is a k-times} CNS order if there are at least k pairwise Z^- inequivalent CNS bases inO. Theorem 1.1 gives the following.

Corollary 2.1. LetK be an algebraic number field of degree 3. Then there are at most finitely many three times CNS orders inK.

3. Results over finitely generated domains

Let A be a domain with quotient field L of characteristic 0. Suppose that A is integrally closed, and thatAis finitely generated overZ^{as a}Z-algebra. LetKbe a finite extension of Lof degree at least 3,AK the integral closure of AinK, andO an A-order inK, that is a subring of A_K which contains Aand which has quotient fieldK. Consider the equation

A[↵] =O ^{in ↵}2O^{. (3.1)}

The solutions of this equation can be divided into A-equivalence classes, where two elements↵, ofO^{are called}A-equivalentif =u↵+afor somea2 Aand u 2 A^⇤. HereA^⇤denotes the multiplicative group of invertible elements of A. As is known (see Roquette [19]), A^⇤is finitely generated.

It was proved by Gy˝ory [10] that the set of↵ with (3.1) is a union of finitely many A-equivalence classes. An explicit upper bound for the number of these A- equivalence classes has been derived by Evertse and Gy˝ory [5]. An effective version has been established by Gy˝ory for certain special types of domains [11].

We now formulate our generalizations of the results from the previous sections to A-orders. We call anA-orderO^ktimes monogenic,if equation (3.1) has at least k A-equivalence classes of solutions.

Theorem 3.1. Let Abe a domain with quotient fieldLof characteristic0which is integrally closed and finitely generated overZ^{, and letK} be a finite extension ofL of degree 3. Then there are at most finitely many three times monogenicA-orders in K.

We now turn to two times monogenic A-orders. Let again K be a finite ex- tension ofL of degree at least 3. We callO^{anA-order in K oftype I}if there are

↵, 2O^{and a}a¹3^aa²4 2GL(2,L)such that

K =L(↵), O= A[↵] = A[ ], = a1↵+a2

a₃↵+a₄, a36=0. (3.2) It should be noted that in the previous section (with L = Q^{, A} = Z) we had in our definition (1.2) of orders of type I the stronger requirement ^a_a¹₃^a_a²₄ 2GL(2,Z⁾ instead of ^aa¹3^aa²4 2 GL(2,Q). In fact, if Ais a principal ideal domain, we can choosea1,a2,a3,a4 in (3.2) such thata1,a2,a3,a4 2 Aand the ideal generated bya1, . . . ,a4equals A. In that case, according to Lemma 6.4 proved in Section 6 below, (3.2) implies that ^a_a¹₃^a_a²₄ 2GL(2,A).

A-orders of type II exist only in extensions of L of degree 4. Thus, letK be an extension of L of degree 4. We call O^{an A-order in K oftype II}if there are

↵, 2O^anda0,a1,a2,b0,b1,b22 Awitha0b06=0, such that

K =L(↵), O= A[↵] =A[ ], (3.3)

=a₀↵²+a₁↵+a₂, ↵=b₀ ²+b₁↵+b₂.

Theorem 3.2. Let Abe a domain with quotient fieldLof characteristic0which is integrally closed and finitely generated overZ^{, and letK}be a finite extension ofL.

Denote byGthe normal closure ofKoverL.

(i) Suppose [K : L] = 3. Then every two times monogenic A-order inK is of type I.

(ii) Suppose[K : L] =4andGal(G/L)⇠= S4. Then there are only finitely many two times monogenic A-orders inK which are not of type I or type II.

(iii) Suppose[K : L] 5and thatK is four times transitive overL. Then there are only finitely many two times monogenic A-orders in K which are not of type I.

4. Equations with unknowns from a finitely generated multiplicative group

The main tools in the proofs of Theorems 3.1 and 3.2 are finiteness results on poly- nomial equations of which the unknowns are taken from finitely generated multi- plicative groups. In this section, we have collected what is needed. Below, Gis a field of characteristic 0.

Lemma 4.1. Let a1,a2 2 G^⇤ and let0 be a finitely generated subgroup ofG^⇤. Then the equation

a1x1+a2x2=1 inx1,x220 (4.1) has only finitely many solutions.

Proof. See Lang [17].

A pair(a1,a2)2(G^⇤)²=G^⇤⇥G^⇤is callednormalizedif(1,1)is a solution to (4.1),i.e.,a1+a2 = 1. If (4.1) has a solution,(y1,y2), say, then by replacing (a1,a2)by(a1y1,a2y2)we obtain an equation like (4.1) with a normalized pair of coefficients, whose number of solutions is the same as that of the original equation.

Lemma 4.2. Let0be a finitely generated subgroup ofG^⇤. There is a finite set of normalized pairs in (G^⇤)², such that for every normalized pair(a₁,a₂) 2 (G^⇤)² outside this set, equation(4.1)has at most two solutions, the pair(1,1)included.

Proof. This result is due to Evertse, Gy˝ory, Stewart and Tijdeman [6]; see also [12].

We note that the proof depends ultimately on the Subspace Theorem, hence it is ineffective.

We consider more generally polynomial equations

f(x1, . . . ,xn)=0 inx1, . . . ,xn 20 (4.2) where f is a non-zero polynomial fromG[X1, . . . ,Xn]and0is a finitely generated subgroup ofG^⇤. Denote byT an auxiliary variable. A solution(x1, . . . ,xn)of (4.2) is calleddegenerate, if there are integersc1, . . . ,cn, not all zero, such that

f(x1T^c1, . . . ,xnT^cn)⌘0 identically inT, (4.3) andnon-degenerateotherwise.

Lemma 4.3. Let f be a non-zero polynomial fromG[X1, . . . ,Xn]and0a finitely generated subgroup ofG^⇤. Then equation(4.2)has only finitely many non-degen- erate solutions.

Proof. Given a multiplicative Abelian groupH, we denote byHⁿitsn-fold direct product with componentwise multiplication.

LetVbe the hypersurface given by f =0. Notice that the degenerate solutions xare precisely those, for which there exists an algebraic subgroup H of(G^⇤)ⁿ of dimension 1 such thatxH ✓ V. By a theorem of Laurent [18], the intersection V\0ⁿis contained in a finite union of cosetsx1H1[· · ·[xrHrwhereH1, . . . ,Hr

are algebraic subgroups of(G^⇤)ⁿ,x1, . . . ,xrare elements of0ⁿ, andxiHi ✓Vfor i = 1, . . . ,r. The non-degenerate solutions in our lemma are precisely the zero- dimensional cosets amongx1H1, . . . ,xrHr, while the degenerate solutions are in the union of the positive dimensional cosets.

5. Finitely generated domains

We recall some facts about domains finitely generated overZ^.

Let Abe an integrally closed domain with quotient fieldL of characteristic 0 which is finitely generated overZ^{. Then A}is a Noetherian domain. Moreover, A is aKrull domain; seee.g. Bourbaki [3], Chapter VII, Section 1. This means the following. Denote byP^(A)the collection of minimal non-zero prime ideals of A, these are the non-zero prime ideals that do not contain a strictly smaller non-zero prime ideal. Then there exist normalized discrete valuations ordp(p ₂ P^(A))on L, such that the following conditions are satisfied:

for everyx 2K^⇤there are only finitely manyp₂P^(A)with

ord_p(x)6=0, (5.1)

A= x 2K : ordp(x) 0 forp₂P^{(A) , (5.2)}

p₌ x 2 A : ordp(x) >0 forp₂P^{(A). (5.3)}

These valuations ordpare uniquely determined. As is easily seen, forx,y2L^⇤we have

ordp(x)=ordp(y)for allp₂P^(A)()x y ¹2 A^⇤. (5.4) LetGbe a finite extension ofL. Denote by AGthe integral closure of AinG, and by A^⇤_G the unit group, i.e., group of invertible elements of AG. We will apply the results from Section 4 with0= A^⇤_G. To this end, we need the following lemma.

Lemma 5.1. The group A^⇤_Gis finitely generated.

Proof. The domainAGis contained in a freeA-module of rank[G :L]. SinceAis Noetherian, the domainAGis finitely generated as anA-module, and so it is finitely generated as an algebra overZ. Then by a theorem of Roquette [19], the groupA^⇤_G is finitely generated.

6. Other auxiliary results

We have collected some elementary lemmas needed in the proofs of Theorems 3.1 and 3.2. Let Abe an integrally closed domain with quotient fieldLof characteristic 0 which is finitely generated overZ^{, andK} a finite extension ofLwith[K :L] =:

d 3. Denote byG the normal closure of K over L. Let ₁ = id, . . . , _d be the distinct L-isomorphisms of K in G, and for ↵ 2 K write ↵⁽ⁱ⁾ := i(↵)for i = 1, . . . ,d. Denote by AK and AG the integral closures of A in K and G, respectively, and by A^⇤_Gthe multiplicative group of invertible elements of AG.

Thediscriminantof↵2K is given by DK/L(↵):= Y

1i<jd

⇣

↵⁽ⁱ⁾ ↵^(j)⌘2

.

This is an element ofL. We have L(↵)= K if and only if all conjugates of↵are distinct, hence if and only if DK/L(↵) 6= 0. Further, if↵ is integral over Athen

D_K/L(↵)2 Asince Ais integrally closed.

Lemma 6.1. Let↵, 2 A_K and suppose that L(↵)= L( )= K, A[↵] = A[ ]. Then

(i)

(i) (j)

↵⁽ⁱ⁾ ↵^(j) 2 A^⇤_G for i,j 2{1, . . . ,d},i 6= j, (ii) D_K/L( )

DK/L(↵) 2 A^⇤.

Proof. (i) Leti,j 2 {1, . . . ,d},i 6= j. We have = f(↵)for some f 2 A[X]. Hence

(i) (j)

↵⁽ⁱ⁾ ↵^(j) = f(↵⁽ⁱ⁾) f(↵^(j))

↵⁽ⁱ⁾ ↵^(j) 2 AG.

Likewise(↵⁽ⁱ⁾ ↵^(j))/( ^{(i) (j)})2 AG. This proves (i).

(ii) We have on the one hand, DK/L( )/DK/L(↵)2L^⇤, on the other hand DK/L( )

DK/L(↵) = Y

1i<jd

(i) (j)

↵⁽ⁱ⁾ ↵^(j)

!2

2A^⇤_G.

Since Ais integrally closed, this proves (ii).

We call two elements ↵, of K L-equivalent if = u↵ +a for someu 2 L^⇤,a2L.

Lemma 6.2. Let↵, 2 AK and suppose that L(↵)= L( )= K, A[↵] = A[ ], and↵, areL-equivalent. Then↵, are A-equivalent.

Proof. By assumption, = u↵+awithu 2 L^⇤,a 2 L. By the previous lemma, u^{d(d 1)}= DK/L( )/DK/L(↵)2 A^⇤, and thenu2 A^⇤since Ais integrally closed.

Consequently,a = u↵is integral over A. Hencea 2 A. This shows that↵, are A-equivalent.

For↵2KwithK = L(↵)we define the ordered(d 2)-tuple

⌧(↵):=

⇣↵⁽³⁾ ↵⁽¹⁾

↵⁽²⁾ ↵⁽¹⁾, . . . ,↵^(d) ↵⁽¹⁾

↵⁽²⁾ ↵⁽¹⁾

⌘

. (6.1)

Lemma 6.3. (i)Let↵, with L(↵) = L( ) = K. Then↵, are L-equivalent if and only if⌧(↵)=⌧( ).

(ii)Let↵, 2 A_K and suppose thatL(↵)= L( )= K, A[↵] = A[ ]. Then↵, are A-equivalent if and only if⌧(↵)=⌧( ).

Proof. (i) If↵, are L-equivalent, then clearly⌧(↵) =⌧( ). Assume conversely

that⌧(↵)=⌧( ). Then there are uniqueu2G^⇤,a2G such that

( ⁽¹⁾, . . . , ^(d))=u(↵⁽¹⁾, . . . ,↵^(d))+a(1, . . . ,1). (6.2) In fact, the unicity of u,a follows since thanks to our assumption K = L(↵), the numbers↵⁽¹⁾, . . . ,↵^(d) are distinct. As for the existence, observe that (6.2) is satisfied withu=( ^{(2) (1)})/(↵⁽²⁾ ↵⁽¹⁾),a= ⁽¹⁾ u↵⁽¹⁾.

Take from the Galois group Gal(G/L). Then ₁, . . . , _d is a per- mutation of theL-isomorphisms 1, . . . , _d : K ,!G. It follows that permutes (↵⁽¹⁾, . . . ,↵^(d))and( ⁽¹⁾, . . . , ^(d))in the same way. So by applying to (6.2) we obtain

( ⁽¹⁾, . . . , ^(d))= (u)(↵⁽¹⁾, . . . ,↵^(d))+ (a)(1, . . . ,1).

By the unicity ofu,a in (6.2) this implies (u) = u, (a) = a. This holds for every 2Gal(G/L). So in factu2L^⇤,a2L, that is,↵, areL-equivalent.

(ii) Use Lemma 6.2.

We denote by(a1, . . . ,ar)the ideal ofAgenerated bya1, . . . ,ar.

Lemma 6.4. Let↵, 2 A_K withL(↵)=L( )= K,A[↵] = A[ ]. Suppose there is a matrix ^a_a¹₃^a_a²₄ 2GL(2,L)with

= a1↵+a2

a₃↵+a₄, a36=0, (6.3)

a1,a2,a3,a42 A, (a1,a2,a3,a4)=(1). (6.4) Then ^a_a¹₃^a_a²₄ 2GL(2,A).

Remark. LetO^{be an} A-order of type I, as defined in Section 3. Then there exist

↵, withO= A[↵] = A[ ], and a matrixU := ^aa¹3^aa²4 2GL(2,L)with (6.3). If Ais a principal ideal domain then by taking a suitable scalar multiple ofU we can arrange that (6.4) also holds, and thus, thatU 2GL(2,A).

Proof. Since↵2AKandL(↵)=K, it has a monic minimal polynomial f 2 A[X] over Lof degreed. Moreover, sinceA[ ] = A[↵], we have

=r₀+r₁↵+ · · · +r_{d 1}↵^{d 1}withr₀, . . . ,r_{d 1}2 A. (6.5) Hence

(a3X+a4)(rd 1X^{d 1}+ · · · +r0) a1X a2=a3rd 1f(X). (6.6) Equating the coefficients, we see that

a4r0 a22a3rd 1A, a4r1+a3r0 a12a3rd 1A, (6.7) a4rj +a3rj 12a3rd 1A(j =2, . . . ,d 1). (6.8)

We first prove that

a₃^{1 j}rj 2 A for j =1, . . . ,d 1. (6.9) In fact, we prove by induction oni (1i d 1), the assertion thata₃^{1 j}r_j 2 A for j = 1, . . . ,i, anda₃^{1 i}rj 2 Afor j = i +1, . . . ,d 1. For i = 1 this is clear. Let 2  i  d 1 and suppose that the assertion is true fori 1 instead of i. Thensj := a²₃ ⁱrj 2 Afor j = i, . . . ,d 1. Further, by (6.8), we have a4sj +a3sj 1 = zja3sd 1withzj 2 Afor j = i, . . . ,d 1. Next, by (6.7) we havea1,a22(a3,a4), and then(a3,a4)=(1)by (6.4). That is, there arex,y 2 A withxa3+ya4 =1. Consequently, for j =i, . . . ,d 1, we have

s_j =(xa₃+ya₄)s_j =a₃(xs_j +y(z_js_{d 1} s_{j 1}))2a₃A,

i.e.,a₃^{1 i}r_j = a₃¹s_j 2 A. This completes our induction step, and completes the proof of (6.9).

Now define the binary formF(X,Y):=Y^df(X/Y). Then (6.6) implies a3rd 1F(X,Y)=(a3X+a4Y)(· · ·) Y^{d 1}(a1X+a2Y).

SubstitutingX =a₄, Y = a₃, and using (6.9), it follows that

F(a4, a3)=s ¹(a1a4 a2a3)withs 2 A. (6.10) Denote by↵⁽¹⁾, . . . ,↵^(d)the conjugates of↵, and by ⁽¹⁾, . . . , ^(d)the correspond- ing conjugates of . Then for the discriminant of we have, by (6.3), (6.10),

DK/L( )= Y

1i<jd

( ^{(i) (j)})²

=(a1a4 a2a3)^{d(d 1)} Yd

i=1

(a4+a3↵⁽ⁱ⁾)

! 2d+2

Y

1i<jd

(↵⁽ⁱ⁾ ↵^(j))²

=(a₁a₄ a₂a₃)^{d(d 1)}F(a₄, a₃) ^2d+2D_K/L(↵)

=s^{2d 2}(a₁a₄ a₂a₃)^{(d 1)(d 2)}D_K/L(↵).

On the other hand, by Lemma 6.1, (ii) we have D_K/L( )/D_K/L(↵) 2 A^⇤. Using also that Ais integrally closed, it follows thata₁a₄ a₂a₃ 2 A^⇤. This completes our proof.

7. Proof of Theorem 3.1

The proof splits into two parts. Consider 2 AK with K = L( ). The first part, which is Lemma 7.1 below, implies that the set of such that A[ ] is three times monogenic, is contained in a union of at most finitely many L-equivalence

classes. The second part, which is Lemma 7.2 below, implies that ifC^{is a givenL-} equivalence class, then the set of 2C^{such thatA}[ ]is two times monogenic, is in a union of at most finitely many A-equivalence classes. (Lemma 7.2 is used in the proof of Theorem 3.2 as well, therefore it deals with two times monogenic orders.) Any three times monogenic A-order inKcan be expressed asA[ ]. A combination of Lemmas 7.1 and 7.2 clearly yields that the set of such lies in finitely many A- equivalence classes. Since A-equivalent give rise to equal A-orders A[ ], there are only finitely many three times monogenic orders inK.

Lemma 7.1. The set of such that

2AK, L( )=K, A[ ]is three times monogenic (7.1) is contained in a union of at most finitely manyL-equivalence classes.

Proof. Assume the contrary. Then there is an infinite sequence of triples {( _1p, _2p, _3p): p=1,2, . . .}such that

hp2 AK, L( hp)= Kforh=1,2,3, p=1,2, . . .; (7.2)

1p(p=1,2, . . .)lie in differentL-equivalence classes (7.3) and for p=1,2, . . .,

⇢A[ 1p] = A[ 2p] = A[ 3p],

1p, _2p, _3plie in different A-equivalence classes (7.4) (so the 1pplay the role of in the statement of our lemma). For any three distinct indicesi,j,kfrom{1, . . . ,d}, and forh=1,2,3,p=1,2, . . ., put

(i jk)

hp :=

(i) hp

(j) hp (i) hp

(k) hp

.

By (7.2), these numbers are well-defined and non-zero.

We start with some observations. Leti, j,kbe any three distinct indices from {1, . . . ,d}. By Lemma 6.1 and the obvious identities _hp^{(i jk)} + _hp^{(k ji)} = 1 (h = 1,2,3), the pairs( _hp^{(i jk)}/ ₁^{(i jk)}_p , _hp^{(k ji)}/ _1p^{(k ji)})(h=1,2,3) are solutions to

(i jk)

1p x + ₁^{(k ji)}_p y =1 inx,y2 A^⇤_G. (7.5)

Notice that (7.5) has solution (1,1). So according to Lemmas 4.2, 5.1, there is a finite set Ai jk such that if ₁^{(i jk)}_p 62 Ai jk, then (7.5) has at most two solu- tions, including (1,1). In particular, there are at most two distinct pairs among ( _hp^{(i jk)}/ _1p^{(i jk)}, _hp^{(k ji)}/ ₁^{(k ji)}_p ) (h=1,2,3). Consequently,

(i jk)

1p 62Ai jk =)two among ₁^{(i jk)}_p , ₂^{(i jk)}_p , ₃^{(i jk)}_p are equal. (7.6)

We start with the cased=3. Then⌧( _hp)=( _hp⁽¹³²⁾)forh=1,2,3. By (7.3) and Lemma 6.3,(i) the numbers _1p⁽¹³²⁾ (p = 1,2, . . .) are pairwise distinct. By (7.6) and Lemma 6.3,(ii), for all but finitely many p, two among the numbers _hp⁽¹³²⁾ (h = 1,2,3) are equal and hence two among hp (h = 1,2,3) are A-equivalent which contradicts (7.4).

Now assume d 4. We have to distinguish between subsets {i,j,k} of {1, . . . ,d}and indiceshfor which there are infinitely many distinct numbers among

(i jk)

hp (p=1,2, . . .), and{i,j,k}andhfor which among these numbers there are only finitely many distinct ones. This does not depend on the choice of ordering ofi, j,k, since any permutation of(i,j,k)transforms _hp^{(i jk)}into one of( _hp^{(i jk)}) ¹, 1 _hp^{(i jk)},(1 _hp^{(i jk)}) ¹, 1 ( _hp^{(i jk)}) ¹,(1 ( _hp^{(i jk)}) ¹) ¹.

By assumption (7.3) and Lemma 6.3,(i), the tuples⌧( _1p)=( ₁⁽¹³²⁾_p , . . . , _1p^(1d2)) (p = 1,2, . . .) are all distinct. Hence one of the sets {1,2,3}, . . . ,{1,2,d}, say {i, j,k}, is such that there are infinitely many distinct numbers among _1p^{(i jk)} (p= 1,2, . . .). Choose an infinite subsequence of indicespsuch that the numbers _1p^{(i jk)} are pairwise distinct. Suppose there is another subset{i⁰,j⁰,k⁰}6= {i,j,k}such that ifpruns through the infinite subsequence just chosen, then ₁⁽ⁱ_p^0j0k0)runs through an infinite set. Then for some infinite subsequence of these p, the numbers _1p^(i0j0k0) are pairwise distinct. Continuing in this way, we infer that there is a non-empty collectionS of 3-element subsets{i,j,k}of{1, . . . ,d}, and an infinite sequence P ^{of indices} p, such that for each{i,j,k} 2 S the numbers ₁^{(i jk)}_p (p 2 P^{) are} pairwise distinct, while for each{i,j,k}62 S, there are only finitely many distinct elements among _1p^{(i jk)}(p2P^).

Notice that if{i,j,k}62S, then among the equations (7.5) with p2P^{, there} are only finitely many distinct ones, and by Lemmas 4.1, 5.1, each of these equa- tions has only finitely many solutions. Therefore, there are only finitely many dis- tinct numbers among _hp^{(i jk)}/ _1p^{(i jk)}hence only finitely many among _hp^{(i jk)}(h=2,3, p 2P). Conversely, if{i, j,k} 2S^,h 2{2,3}, there are infinitely many distinct numbers among _hp^{(i jk)}(p2P). For if not, then by the same argument, interchang- ing the roles of hp, 1p, it would follow that there are only finitely many distinct numbers among _1p^{(i jk)}(p2P), contradicting{i, j,k}2S^.

We conclude that there is an infinite subsequence of p, which after renaming we may assume to be 1,2, . . ., such that forh=1,2,3,

(i jk)

hp (p=1,2, . . .)are pairwise distinct if{i,j,k}2S^{, (7.7)} there are only finitely many distinct numbers among

(i jk)

hp (p=1,2, . . .)if{i,j,k}62S^{. (7.8)}

Notice that this characterization ofSis symmetric in hp(h=1,2,3); this will be used frequently.

We frequently use the following property ofS^{: ifi, j,k,l}are any four distinct indices from{1, . . . ,d}, then

{i,j,k}2S =){i,j,l}2S^or{i,k,l}2S^{. (7.9)} Indeed, if{i, j,l},{i,k,l}62S ^{then also}{i, j,k}62S^since _hp^{(i jk)}= hp^{(i jl)}/ _hp^(ikl).

Pick a set from S, which without loss of generality we may assume to be {1,2,3}. By (7.9), for k = 4, . . . ,d at least one of the sets {1,2,k}, {1,3,k} belongs toS. Define the set of pairs

C:=n

(j,k): j 2{2,3},k2{3, . . . ,d}, j 6=k,{1,j,k}2S^{o. (7.10)} Thus, for eachk 2 {3, . . . ,d}there is j with(j,k) 2 C. Further, for every p = 1,2, . . .there is a pair(j,k)2C^{such that}

(1jk)

1p 6= _2p^(1jk).

Indeed, if this were not the case, then since _hp^(12k) = hp^(13k) (123)

hp , it would follow that for some p,

(12k)

1p = _2p^(12k)fork=3, . . . ,d,

and then ⌧( _1p) = ⌧( _2p). Together with Lemma 6.3,(ii) this would imply that

1p, 2p are A-equivalent, contrary to (7.4). Clearly, there is a pair (j,k) 2 C such that _1p^(1jk) 6= _2p^(1jk) for infinitely many p. After interchanging the indices 2 and 3 if j = 3 and then permuting the indices 3, . . . ,d, which does not affect the above argument, we may assume that j =2,k = 3. That is, we may assume that {1,2,3}2S^and

(123)

1p 6= ₂⁽¹²³⁾_p for infinitely many p.

We now bring (7.6) into play. It implies that for infinitely many pwe have _3p⁽¹²³⁾ 2 { _1p⁽¹²³⁾, ₂⁽¹²³⁾_p }. After interchanging 1p, 2p(which does not affect the definition ofSor the above arguments) we may assume that{1,2,3}2S^and

(123)

1p = ₃⁽¹²³⁾_p 6= ₂⁽¹²³⁾_p (7.11)

for infinitely many p.

We repeat the above argument. After renaming again, we may assume that the above infinite sequence of indices pfor which (7.11) is true is p = 1,2, . . ., and thus, (7.7) and (7.8) are true again. Define again the setCby (7.10). Similarly as above, we conclude that there is a pair(j,k) 2 C such that among p = 1,2, . . . there is an infinite subset with ₁⁽¹_p^jk) 6= _3p^(1jk). Then necessarily,k6=3. After inter- changing 2 and 3 if j =3 (which does not affect (7.11)) and rearranging the other

indices 4, . . . ,d, we may assume that j = 2,k = 4. Thus,{1,2,3},{1,2,4}2S and there are infinitely many pfor which we have (7.11) and

(124)

1p 6= _3p⁽¹²⁴⁾.

By (7.6), for all but finitely many of these pwe have ₂⁽¹²⁴⁾_p 2{ _1p⁽¹²⁴⁾, ₃⁽¹²⁴⁾_p }. After interchanging 1p, _3pif necessary, which does not affect (7.11), we may conclude that{1,2,3},{1,2,4}2Sand there are infinitely many pwith (7.11) and

(124)

1p = ₂⁽¹²⁴⁾_p 6= _3p⁽¹²⁴⁾. (7.12)

Next, by (7.9), at least one of {1,3,4}, {2,3,4} belongs toS. Relations (7.11), (7.12) remain unaffected if we interchange _hp⁽¹⁾ and _hp⁽²⁾, so without loss of gen- erality, we may assume that {1,3,4} 2 S. By (7.6), for all but finitely many of the pwith (7.11) and (7.12), at least two among the numbers _hp⁽¹³⁴⁾ (h = 1,2,3) must be equal. Using (7.11), (7.12) and _hp⁽¹³⁴⁾ = _hp⁽¹²⁴⁾/ _hp⁽¹²³⁾, it follows that {1,2,3},{1,2,4},{1,3,4} 2 S and for infinitely many pwe have (7.11), (7.12)

and (134)

2p = ₃⁽¹³⁴⁾_p 6= _1p⁽¹³⁴⁾. (7.13)

We now show that this is impossible. For convenience we introduce the notation

˜_hp⁽ⁱ⁾ :=

(i) hp

(4) hp (3) hp

(4) hp

= _hp⁽⁴ⁱ³⁾

forh = 1,2,3,i = 1,2,3,4, p = 1,2, . . .. Notice that ˜_hp⁽³⁾ = 1, ˜_hp⁽⁴⁾ = 0, and

(i jk)

hp = ^˜

(i) hp ˜_hp^(j)

˜_hp⁽ⁱ⁾ ˜_hp^(k) for any distincti,j,k2{1,2,3,4}. Thus, (7.11)–(7.13) translate into

˜_1p⁽¹⁾ ˜_1p⁽²⁾

˜_1p⁽¹⁾ 1 =

˜_3p⁽¹⁾ ˜_3p⁽²⁾

˜_3p⁽¹⁾ 1 6=

˜₂⁽¹⁾_p ˜_2p⁽²⁾

˜₂⁽¹⁾_p 1 , (7.14)

˜_1p⁽¹⁾ ˜_1p⁽²⁾

˜_1p⁽¹⁾ = ˜_2p⁽¹⁾ ˜_2p⁽²⁾

˜₂⁽¹⁾_p 6= ˜₃⁽¹⁾_p ˜_3p⁽²⁾

˜_3p⁽¹⁾ , (7.15)

˜_2p⁽¹⁾ 1

˜_2p⁽¹⁾ =

˜₃⁽¹⁾_p 1

˜_3p⁽¹⁾ 6=

˜_1p⁽¹⁾ 1

˜₁⁽¹⁾_p . (7.16)

We distinguish between the cases{2,3,4}2S^and{2,3,4}62S^.

First suppose that{2,3,4} 2 S. Then by (7.6), there are infinitely many p such that (7.14)–(7.16) hold and at least two among ˜_hp⁽²⁾= _hp⁽⁴²³⁾ (h=1,2,3) are